这是我能想到的最好的算法。
def get_primes(n):
numbers = set(range(n, 1, -1))
primes = []
while numbers:
p = numbers.pop()
primes.append(p)
numbers.difference_update(set(range(p*2, n+1, p)))
return primes
>>> timeit.Timer(stmt='get_primes.get_primes(1000000)', setup='import get_primes').timeit(1)
1.1499958793645562
还能做得更快吗?
这段代码有一个缺陷:由于numbers是一个无序集，不能保证numbers.pop()将从集合中移除最低的数字。尽管如此，它还是适用于(至少对我来说)一些输入数字:
>>> sum(get_primes(2000000))
142913828922L
#That's the correct sum of all numbers below 2 million
>>> 529 in get_primes(1000)
False
>>> 529 in get_primes(530)
True
这里是最快的函数之一的两个更新版本(纯Python 3.6)，
from itertools import compress
def rwh_primes1v1(n):
""" Returns a list of primes < n for n > 2 """
sieve = bytearray([True]) * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = bytearray((n-i*i-1)//(2*i)+1)
return [2,*compress(range(3,n,2), sieve[1:])]
def rwh_primes1v2(n):
""" Returns a list of primes < n for n > 2 """
sieve = bytearray([True]) * (n//2+1)
for i in range(1,int(n**0.5)//2+1):
if sieve[i]:
sieve[2*i*(i+1)::2*i+1] = bytearray((n//2-2*i*(i+1))//(2*i+1)+1)
return [2,*compress(range(3,n,2), sieve[1:])]
我可能迟到了，但必须为此添加自己的代码。它使用大约n/2的空间，因为我们不需要存储偶数，我还使用bitarray python模块，进一步大幅减少内存消耗，并允许计算所有高达1,000,000,000的质数
from bitarray import bitarray
def primes_to(n):
size = n//2
sieve = bitarray(size)
sieve.setall(1)
limit = int(n**0.5)
for i in range(1,limit):
if sieve[i]:
val = 2*i+1
sieve[(i+i*val)::val] = 0
return [2] + [2*i+1 for i, v in enumerate(sieve) if v and i > 0]
python -m timeit -n10 -s "import euler" "euler.primes_to(1000000000)"
10 loops, best of 3: 46.5 sec per loop
这是在64bit 2.4GHZ MAC OSX 10.8.3上运行的
这是使用存储列表查找质数的一种优雅而简单的解决方案。从4个变量开始，你只需要测试除数的奇数质数，你只需要测试你要测试的质数的一半(测试9,11,13是否能整除17没有意义)。它将先前存储的质数作为除数进行测试。
# Program to calculate Primes
primes = [1,3,5,7]
for n in range(9,100000,2):
for x in range(1,(len(primes)/2)):
if n % primes[x] == 0:
break
else:
primes.append(n)
print primes
下面是Eratosthenes的一个numpy版本，具有良好的复杂度(低于排序长度为n的数组)和向量化。与@unutbu相比，用46微秒就可以找到100万以下的所有质数。
import numpy as np
def generate_primes(n):
is_prime = np.ones(n+1,dtype=bool)
is_prime[0:2] = False
for i in range(int(n**0.5)+1):
if is_prime[i]:
is_prime[i**2::i]=False
return np.where(is_prime)[0]
计时:
import time
for i in range(2,10):
timer =time.time()
generate_primes(10**i)
print('n = 10^',i,' time =', round(time.time()-timer,6))
>> n = 10^ 2 time = 5.6e-05
>> n = 10^ 3 time = 6.4e-05
>> n = 10^ 4 time = 0.000114
>> n = 10^ 5 time = 0.000593
>> n = 10^ 6 time = 0.00467
>> n = 10^ 7 time = 0.177758
>> n = 10^ 8 time = 1.701312
>> n = 10^ 9 time = 19.322478
使用Sundaram的Sieve，我想我打破了pure-Python的记录:
def sundaram3(max_n):
numbers = range(3, max_n+1, 2)
half = (max_n)//2
initial = 4
for step in xrange(3, max_n+1, 2):
for i in xrange(initial, half, step):
numbers[i-1] = 0
initial += 2*(step+1)
if initial > half:
return [2] + filter(None, numbers)
Comparasion:
C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.get_primes_erat(1000000)"
10 loops, best of 3: 710 msec per loop
C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.daniel_sieve_2(1000000)"
10 loops, best of 3: 435 msec per loop
C:\USERS>python -m timeit -n10 -s "import get_primes" "get_primes.sundaram3(1000000)"
10 loops, best of 3: 327 msec per loop
